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Difficulty: MediumGalvanic Cells and Standard Electrode Potentials
Consider a standard galvanic cell constructed using cobalt and silver half-cells under standard conditions:
Co(aq)2++2e−→Co(s)E∘=−0.28 V\text{Co}^{2+}_{\text{(aq)}} + 2\text{e}^- \rightarrow \text{Co}_{\text{(s)}} \quad E^\circ = -0.28\text{ V}
Ag(aq)++e−→Ag(s)E∘=+0.80 V\text{Ag}^+_{\text{(aq)}} + \text{e}^- \rightarrow \text{Ag}_{\text{(s)}} \quad E^\circ = +0.80\text{ V}
Which of the following statements correctly describes the operational mechanics of this cell?
  1. A
    Cobalt acts as the cathode and undergoes oxidation, leading to a decrease in electrode mass.
  2. Electrons flow spontaneously from the cobalt electrode to the silver electrode through the external conductor.Answer
  3. C
    The standard electromotive force (Ecell∘E^\circ_{\text{cell}}) generated by the cell is +0.52 V+0.52\text{ V}.
  4. D
    Nitrate ions (NO3−\text{NO}_3^-) from the salt bridge migrate into the silver half-cell to maintain electrical neutrality.

Answer

Electrons flow spontaneously from the cobalt electrode to the silver electrode through the external conductor.
In a galvanic cell, oxidation occurs at the electrode with the lower standard reduction potential (cobalt electrode, making it the anode). Reduction occurs at the electrode with the higher standard reduction potential (silver electrode, making it the cathode). Electrons are generated at the anode by oxidation and travel through the external circuit to the cathode.

Step-by-Step Solution

1
Identify the anode and cathode based on standard reduction potentials (E∘E^\circ).
The half-cell with the more negative standard reduction potential (E∘(Co2+/Co)=−0.28 VE^\circ(\text{Co}^{2+}/\text{Co}) = -0.28\text{ V}) undergoes oxidation at the anode. The half-cell with the more positive potential (E∘(Ag+/Ag)=+0.80 VE^\circ(\text{Ag}^+/\text{Ag}) = +0.80\text{ V}) undergoes reduction at the cathode.
Species with higher standard reduction potentials are more easily reduced, while those with lower reduction potentials are more easily oxidized.
2
Determine the direction of electron flow in the external circuit.
Oxidation at the cobalt anode releases electrons: Co(s)→Co(aq)2++2e−\text{Co}_{\text{(s)}} \rightarrow \text{Co}^{2+}_{\text{(aq)}} + 2\text{e}^-. These electrons flow through the wire toward the silver cathode.
Electrons always flow spontaneously from the anode (site of oxidation) to the cathode (site of reduction) in a galvanic cell.
3
Calculate the cell potential (Ecell∘E^\circ_{\text{cell}}) to verify consistency.
Ecell∘=Ecathode∘−Eanode∘=+0.80 V−(−0.28 V)=+1.08 VE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.80\text{ V} - (-0.28\text{ V}) = +1.08\text{ V}.
A positive standard cell electromotive force confirms that the cell reaction is spontaneous in the stated direction.

Key Concept

Galvanic Cell Mechanics and Standard Electrode Potentials
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Topics

Galvanic Cells and Standard Electrode Potentials